Hi Katie,
This is probably a bit late given the date of your post, but I was having
similar problems with my own work and thought I'd respond anyway. I'm not
sure that the script you've written here will fit 3D vectors for your 3D
nmds. I tried it and it doesn't seem to work for me - it only gives
Does censReg expect from panel data to be balanced? Because in my case it is
unbalanced. Could this be a reason for errors?
Best,
Igors
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any ideas?
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__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listi
I assume you mean Cohen's kappa. This is not what the OP is asking about. The
OP wants to know how to test for a difference in the proportions of 1's.
Cohen's kappa will tell you what the level of agreement is between the two
tests. This is something different.
Also, the OP has now clarified th
Dear list members,
I use the package inlinedocs to create documentation for a package that
I am building. The function package.skeleton.dx() is used to convert the
source *.R file into the *.Rd documentation file. Usually this works
like a charm, but on several occasions I got messages like t
Thanks, Duncan, for your reply. You are right, the () in my code are not
correct. Maybe my problem is that I do not really understand the exact
effect of this dot . I have no tried with the following file in my /R
folder in the package:
mainfunction<- function(x) {
x2 <- .subfunction1(x)
Seems as whether I found a (clumsy) workaround:
1. options(warn =2)
2. run package.skeleton.dx, now an error is produced with the name of
the file package.skeleton was actually working on
3. The number of the erroneous input string should correspond to the
line number in that file.
Just in ca
On 11-09-07 6:25 PM, Mike Miller wrote:
I'm getting the impression from on-line docs that R cannot work with
single-precision floating-point numbers, but that it has a pseudo-mode for
single precision for communication with external programs.
I don't mind that R is using doubles internally, but
On 11-09-07 6:07 PM, Dale Coons wrote:
Doing a visual graphic and trying to make it "pretty"
Here's simple chart to play with:
library(rgl)
dotframe<-data.frame(x=c(0,7,0,0,-7,0),y=c(0,0,7,0,0,-7),z=c(7,0,0,-7,0,0))
dotframe plot3d(dotframe$x,dotframe$y,dotframe$z, radius=3,
type='s',col=c('red
Hi Andrew,
maybe history() helps you?
It also allows pattern search (using grep internally).
hth.
Am 08.09.2011 07:03, schrieb andrewH:
> Is there any way to search the console during an interactive session? I've
> looked and looked, and can not find one. In some add-on package, maybe?
>
> Sorr
Hi
Beware of shooting in your leg. Instead of poisoning your workspace with
numerous objects called "obj.n" or something like that you help yourself
with creating single object of type list. Parts of a list can be accessed
much more easily than this kind of get construction.
I use R quite a lo
On 11-09-08 5:28 AM, Jannis wrote:
Thanks, Duncan, for your reply. You are right, the () in my code are not
correct. Maybe my problem is that I do not really understand the exact
effect of this dot . I have no tried with the following file in my /R
folder in the package:
mainfunction<- function
On Sep 8, 2011, at 03:18 , zhenjiang xu wrote:
> Thanks for all your replies. I am using rowsum() and it looks efficient. I
> hope I could do some benchmark sometime in near future and let people know.
> Or is there any benchmark result available?
I'm a bit surprised that no-one thought of xtabs
On 09/08/2011 12:02 AM, dadrivr wrote:
Hi,
I'm trying to reshape my data set from wide to tall format for multilevel
modeling. Unfortunately, the function I typically use (make.univ from the
multilevel package) does not appear to work with unbalanced data frames,
which is what I'm dealing with.
Hi Steve,
Thanks for the reply. I am a complete novice to R and to using SVM as well
and therefore the terms that I'd used maybe non standard(I am not sure how
to call them technically). Nevertheless the number of data points I had
mentioned are just instances. But as I mentioned, these are bound
Hi Paul,
I'm struggling to find the R code as well on the books website. Did you
managed to get it and are you willing to share it with me?
BR
Peter
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Hi everyone,
I have a zero vector of length N and I would like to randomly allocate the
value 1 to one of the values of this vector. I presume I have to use the
uniform distribution but could someone tell me how I should process?
Thanks in advance,
Boris
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http://
Hi Boris.
Here is one approach:
N<-100
a<-rep(0,N)
a[sample(N,1)]<-1
a
which(a==1)
Look ?sample, ?which.
Andrija
On Thu, Sep 8, 2011 at 10:42 AM, Boris Beranger wrote:
> Hi everyone,
>
> I have a zero vector of length N and I would like to randomly allocate the
> value 1 to one of the values
On 09/08/2011 04:42 AM, Boris Beranger wrote:
I have a zero vector of length N and I would like to randomly allocate the
value 1 to one of the values of this vector. I presume I have to use the
uniform distribution but could someone tell me how I should process?
rmultinom(m,1,rep(1,N))
where m
If you read the help file on the step() function
?step
you will see a reference to BIC under the description of the k= argument.
This suggests that you could try:
BIC.fitted = step(glm.fit, k=log(dim(dat)[1]))
Jean
Andra Isan wrote on 09/07/2011 06:12:19 PM:
>
> Hi All,
> After fitt
Hi everybody,
Thank you so much for all the explanations. Now it is much more clear to me.
And sorry for the delay in answer but i move office and i was not on line for
some time (not that you are really interested in that, though ;-)).
I should have sent a dummy example, my real data is a lit
Hello.
I found the behavior of knn(
http://stat.ethz.ch/R-manual/R-devel/library/class/html/knn.html) function
looking very strange.
Consider the toy example.
> library(class)
> train <- matrix(nrow=5000,ncol=2,data=rnorm(1,0,1))
> test <- matrix(nrow=10,ncol=2,data=rnorm(20,0,1))
> cl <- rep(
Thank you very much Andrija,
I have been do some research and was about to post the same solution.
Boris
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I'm not at all certain what you wish to search: R objects? Past commands?
If the latter, others have offered suggestions. If the former, what
about simply ls()?
Or if you mean something else entirely, please clarify.
Sarah
On Thu, Sep 8, 2011 at 1:03 AM, andrewH wrote:
> Is there any way to se
Hi
(R 2.13.1, OSX 10.6.8)
I am trying to use predict.rma with continuous and categorical variables. The
argument newmods in predict.rma seems to handle coviariates, but appears to
falter on factors. While I realise that the coefficients for factors provide
the answers, the goal is to eventua
Hello,
I am using cbind in a lm-model. For standard lm-models
the r.squared can be easily extracted with summary(model)$r.squared,
but that is not working in in the case with cbind.
Here an example to illustrate the problem:
a <- c(1,3,5,2,5,3,1,6,7,2,3,2,6)
b <- c(12,15,18,10,18,22,9,7,9,23,12,
It looks like you are building a regression model. With such a large number of
rows, you should try to limit the size of the trees by setting nodesize to
something larger than the default (5). The issue, I suspect, is the fact that
the size of the largest possible tree has about 2*nodesize nod
I have a matrix called mat and y is the column number of my response and
x is a vector of the column numbers of my terms. The variable name of y
can change, so I don't want to hardcode it. I can find out the name as
follows:
>names(mat)[y]
[1] "er12.l"
Then I can run the regression
Hi, anyone has experience with 3D plot (ex: in package RGL) I have a
question, I draw a 3D plot of country, year and sales in z axis but when the
type is "h" then it's ok but when I want to link the points and type is 'l'
lines it's a mess Is there a way to link the points only in one direction?
Fo
On Sep 8, 2011, at 14:53 , Johannes Radinger wrote:
> Hello,
>
> I am using cbind in a lm-model. For standard lm-models
> the r.squared can be easily extracted with summary(model)$r.squared,
> but that is not working in in the case with cbind.
>
> Here an example to illustrate the problem:
> a
I modified an example in the object.size help page to create a function
I want to be able to run:
"mysize" <- function() {
z <- sapply(ls(), function(w) object.size(get(w)))
as.matrix(rev(sort(z))[1:5])
}
mysize()
When I test the lines inside the function it works fine:
> z <- sapply
Hi
>
> Hello,
>
> I am using cbind in a lm-model. For standard lm-models
> the r.squared can be easily extracted with summary(model)$r.squared,
> but that is not working in in the case with cbind.
>
> Here an example to illustrate the problem:
> a <- c(1,3,5,2,5,3,1,6,7,2,3,2,6)
> b <- c(12,15,
On Sep 8, 2011, at 9:03 AM, Bos, Roger wrote:
I have a matrix called mat and y is the column number of my response
and
x is a vector of the column numbers of my terms. The variable name
of y
can change, so I don't want to hardcode it. I can find out the name
as
follows:
name
The answer to my question was :
fmla <- as.formula(names(mat)[y] %+% " ~ .")
mod <- try(rlm(fmla, data=mat[zidx, c(y, x)]), silent=TRUE)
Thanks for your help and the quick response.
Roger
-Original Message-
From: Jean-Christophe BOUËTTÉ [mailto:jcboue...@gmail.com]
Sent: Thursday, Sep
On Sep 8, 2011, at 9:13 AM, Bos, Roger wrote:
I modified an example in the object.size help page to create a
function
I want to be able to run:
"mysize" <- function() {
z <- sapply(ls(), function(w) object.size(get(w)))
as.matrix(rev(sort(z))[1:5])
}
mysize()
When I test the lines ins
Em 8/9/2011 05:24, Viechtbauer Wolfgang (STAT) escreveu:
I assume you mean Cohen's kappa. This is not what the OP is asking
about. The OP wants to know how to test for a difference in the
proportions of 1's. Cohen's kappa will tell you what the level of
agreement is between the two tests. This is
Here is a function I use to look at the sizes of objects:
my.ls <-
function (pos = 1, sorted = FALSE)
{
.result <- sapply(ls(pos = pos, all.names = TRUE), function(..x)
object.size(eval(as.symbol(..x
if (sorted) {
.result <- rev(sort(.result))
}
.ls <- as.data.frame(rbi
francogrex wrote on 09/08/2011 08:08:19 AM:
>
> Hi, anyone has experience with 3D plot (ex: in package RGL) I have a
> question, I draw a 3D plot of country, year and sales in z axis but when
the
> type is "h" then it's ok but when I want to link the points and type is
'l'
> lines it's a mess Is
-- begin included message --
I am trying to create a classification tree using either tree or rpart
but when it comes to plotting the results the formatting I get is
different than what I see in all the tutorials. What I would like to
see is the XX/XX format but all I get is a weird decimal value.
Hi Kristian,
The idea behind projection is that you take an iterate that violates the
constraints and project it onto a point such that it is the nearest point that
satisfies the constraints.
Suppose you have an iterate (w1, w4) that does not satisfy the constraint that
w1 * w4 != (1 + eps)/2.
Hi,
I was trying to overlay/combine two freqpoly plots. The sample code below
illustrates the problem. Essentially, I want to do is:
1. Have the same colour for all the lines in 'Plot 1' (and 'Plot 2').
Currently, all the lines in Plot 1 have different colours and all the lines
in Plot 2 have dif
Dear All,
I would like to minimise a nonlinear function subject to linear inequality
constraints as part of an R program. I have been using the constrOptim
function. I have tried all of the methods that come with Optim, but nothing
finds the correct solution. If I use the correct solution as
have a look at ?formula and the examples
2011/9/8 Bos, Roger :
> I have a matrix called mat and y is the column number of my response and
> x is a vector of the column numbers of my terms. The variable name of y
> can change, so I don't want to hardcode it. I can find out the name as
> follows:
Hi,
I am trying to get the terminal nodes of a plot of a ctree object to look nice.
Using the iris data I have:
library(party)
mtree <- ctree(Species ~ ., data=iris)
plot(mtree,terminal_panel=node_barplot(mtree))
The terminal nodes don't display the species names because the names
are displayed
Hello everyone
I have a couple of questions about the usage of the R function
"read.table(.)". My point of departure is that I want to import a matrix
(consisting of time and daily stock returns of many stocks) in R. Most of
the data is numeric, however some values are missing (blanks) and in o
On Thu, 8 Sep 2011, Duncan Murdoch wrote:
On 11-09-07 6:25 PM, Mike Miller wrote:
I'm getting the impression from on-line docs that R cannot work with
single-precision floating-point numbers, but that it has a pseudo-mode
for single precision for communication with external programs.
I don'
I set maxnodes and nodesize to reasonable levels and everything is
working great now. Thanks for the guidance.
v/r,
John
On Thu, Sep 8, 2011 at 8:58 AM, Liaw, Andy wrote:
> It looks like you are building a regression model. With such a large number
> of rows, you should try to limit the size o
Hi,
I was trying to overlay/combine two freqpoly plots. The sample code below
illustrates the problem. Essentially, I want to do is:
1. Have the same colour for all the lines in 'Plot 1' (and 'Plot 2').
Currently, all the lines in Plot 1 have different colours and all the
lines in Plot 2 have
Hi Brian
On Thu, Sep 8, 2011 at 10:30 AM, Brian Smith wrote:
> Hi,
>
> I was trying to overlay/combine two freqpoly plots. The sample code below
> illustrates the problem. Essentially, I want to do is:
>
> 1. Have the same colour for all the lines in 'Plot 1' (and 'Plot 2').
Then don't map the c
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of jim holtman
> Sent: Thursday, September 08, 2011 6:52 AM
> To: Bos, Roger
> Cc: r-help@r-project.org
> Subject: Re: [R] Variable scoping question
>
> Here is a function I use to
Hello Andrew,
Take a look at the following:
predict(model1, addx=T)
predict(model2, addx=T)
predict(model3, addx=T)
As you can see, the factor was turned into dummy variables. However, the
predict.rma() function does not expand a factor passed via newmods into the
corresponding dummy variables
Hi, I have a vector 'data' of 58 probability values (bounded between 0 and 1)
and want to draw a probability density function of these values. For this, I
used the commands:
data <- runif(58)
a <- density(data, from=0, to=1)
plot(a, type="l",lwd=3)
But then, when I try to approximate the area
I am trying to create a classification tree using either tree or rpart
functions but when it comes to plotting the results the formatting I get is
different than what I see in all the tutorials (like
http://www.youtube.com/watch?v=9XNhqO1bu0A or
http://www.youtube.com/watch?v=m3mLNpeke0I&feature=re
Use gzcon() to make a compressed connection and any
function that write to a connection will write compressed
data. E.g.,
> con <- gzcon(file("tempfile.junk", "wb"))
> x <- as.integer(rep(c(-127, 1, 127), c(3,2,1)))
> writeBin(x, con, size=1)
> close(con)
> q("no")
bill:158% zcat temp
In my.ls() you ought to convert the pos argument
to an environment and consistently use that environment
in the calls to eval, get, and ls in the function.
E.g., with the following modification
my.ls1 <- function (pos = 1, sorted = FALSE, envir = as.environment(pos))
{
.result <- sapply(ls(env
Hi,
If you read carefully the help pages for read.table you get this:
na.stringsa character vector of strings which are to be interpreted as
NA<../../utils/help/NA> values.
Blank fields are also considered to be missing values in logical, integer,
numeric and complex fields.
So, both NAs and bl
On Thu, 8 Sep 2011, William Dunlap wrote:
Use gzcon() to make a compressed connection and any function that write
to a connection will write compressed data. E.g.,
> con <- gzcon(file("tempfile.junk", "wb"))
> x <- as.integer(rep(c(-127, 1, 127), c(3,2,1)))
> writeBin(x, con, size=1)
> cl
Hi,
Use packages "rpart.plot" or "maptree" to enhance the tree drawing.
Or another alternative, use "party" package that offers much more graphing
possibilities.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Thu, Sep 8, 2011 at 5:27 PM, Brian Jensvold wrote:
> I am trying to create a cla
Hi,
Thanks for the reply. For the combined plot, if I use:
ggplot(comb2) + geom_freqpoly(aes(x = value,
y = ..density.., group = X2))
I get the same colour for both the sets of distributions. What I want is one
colour for the first set of distributions, and a different c
Hi,
I think this other post could help:
http://stackoverflow.com/questions/7328104/optim-with-inequality-constraint
Regards,
Carlos Ortega
www.qualityexcellence.es
On Thu, Sep 8, 2011 at 3:54 PM, Liu Evans, Gareth <
gareth.liu-ev...@liverpool.ac.uk> wrote:
> Dear All,
>
> I would like to minimi
Hi,
You can get something better in this way:
plot(mtree,terminal_panel=node_barplot(mtree, ylines=1.2, gap=0.05))
Check node_barplot() for details, although it does not offer for barplots
the possibility to graph it horizontally.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Thu, Sep 8,
On 08/09/2011 11:14 AM, Mike Miller wrote:
On Thu, 8 Sep 2011, Duncan Murdoch wrote:
> On 11-09-07 6:25 PM, Mike Miller wrote:
>
>> I'm getting the impression from on-line docs that R cannot work with
>> single-precision floating-point numbers, but that it has a pseudo-mode
>> for single pre
For bounded density estimation look at the logspline package instead of the
regular density function.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-hel
Look at
area <- sum(a$y)*(a$x[1]-a$y[2])
The problem appears to be "a$x[1]-a$y[2]"; that is not the length of
the base of an approximating rectangle, whatever it is :-)
albyn
On Thu, Sep 08, 2011 at 11:36:23AM -0400, Gonçalo Ferraz wrote:
> Hi, I have a vector 'data' of 58 probability value
On Mon, Sep 5, 2011 at 12:25 AM, Yvonnick Noel wrote:
> Hi Bill,
>
> I once modelled a hysteresis phenomenon (on binary data) with a simple
> logistic model. I am not sure I understand how this pattern appears in your
> data, but in my previous analyses, it appeared as an order effect: The
> resp
I want to plot a multiline annotation including a mathematical expression using
grid graphics (grid.text).
This works fine for a single line, but the math expression is misplaced when I
insert a line break ("\n" escape sequence):
grid.text(expression(paste("Some words here\n more ", sigma)))
Do
I've seen a number of issues with the loading of workspaces discussed
previously, but here's another one... I simply can't load any saved workspace
at all... Here's an example, starting with an empty workspace and creating a
single variable "a".
> a<-1:5
> save.image("a.Rdata")
> rm(a)
> load
Hi All,
I have txt file like :
$ cat data.txt
US 10
UK 12
Ind 4
Germany 14
France 8
> rawdata <- read.table(file='data.txt',sep='\t' , header=FALSE)
> rawdata
V1 V2
1 US 10
2 UK 12
3 Ind 4
4 Germany 14
5 France 8
I want to draw pie chart for the above dat
Mohan L wrote on 09/08/2011 12:35:18 PM:
>
> Hi All,
>
> I have txt file like :
>
> $ cat data.txt
> US 10
> UK 12
> Ind 4
> Germany 14
> France 8
>
> > rawdata <- read.table(file='data.txt',sep='\t' , header=FALSE)
>
> > rawdata
>V1 V2
> 1 US 10
> 2 UK 12
> 3 I
Is your "data" supposed to be observations, or values of the density
of the underlying law?
Also, could you explain the rationale behind :
sum(a$y)*(a$x[1]-a$y[2])
because it is not immediately clear to the reader.
2011/9/8 Gonçalo Ferraz :
> Hi, I have a vector 'data' of 58 probability values (b
Hi,
I wonder if someone can help me. I have built a gam model to predict the
presence of cold water corals and am now trying to evaluate my model by
splitting my dataset into training/test datasets.
In an ideal world I would use the sample() function to randomly select rows of
data for me so fo
I understand this isn't a r specific question. I'm switching departments to
work with the analytics team at my company as a "service" side manager to
better incorporate the analytics process into product design / production.
We're an online gaming company.
As I'm going through tools like R, rapi
My approach has been autodidactic. There's a bookstore near me (Barnes
and Noble at U of Penn) that has a terrific stats section. Amazon also
has a lot of books, including limited online browsing + reader
recommendations.
On 9/8/11, kensuguro wrote:
> I understand this isn't a r specific question
Also, look at ocw.mit.edu for free course notes with statistical content.
On 9/8/11, kensuguro wrote:
> I understand this isn't a r specific question. I'm switching departments to
> work with the analytics team at my company as a "service" side manager to
> better incorporate the analytics proce
Dear Sarah--
I am thinking mainly in terms of long programs run by cut-and-past or some
other batch-like submission, where you can get back a lot of code, some
program outputs, and some error messages, all in a big lump. I want tl look
through that lump to locate all the error or warning messages
Hi,
x <- 0.2*5
is.integer(x)
gives me FALSE because R stores it as a float number, right?
Is there an elegant way to work around that problem? Right now
I'm using
x <- 0.2*5
round(x) == x
which returns TRUE. But more strictly I should use all.equal(),
right?
I somehow just don
"Brian Jensvold" het geskryf
I am trying to create a classification tree using either tree or rpart
but when it comes to plotting the results the formatting I get is
different than what I see in all the tutorials. What I would like to
see is the XX/XX format but all I get is a weird decimal val
Thanks Eik!
I did not know about or remember history. I agree that it solves part of my
problem, but I really want to be able to search my code and the things R has
printed in response as a single block of text. I can cut-and-paste it into a
text editor, but I was hoping that there was a way to do
Hello Alex,
Have you tried the modulus operator?
> 2 %% 1
[1] 0
> 2.1 %% 1
[1] 0.1
~Jason
On 2011.09.08 20:27:14, Alexander Engelhardt wrote:
> Hi,
>
> x <- 0.2*5
> is.integer(x)
>
> gives me FALSE because R stores it as a float number, right?
> Is there an elegant way to work aroun
On Sep 8, 2011, at 19:03 , Albyn Jones wrote:
> Look at
>
>area <- sum(a$y)*(a$x[1]-a$y[2])
>
> The problem appears to be "a$x[1]-a$y[2]"; that is not the length of
> the base of an approximating rectangle, whatever it is :-)
I would assume that that is just a typo for a$x[1]-a$x[2], which
Jeff Newmiller and Dennis,
As always, very helpful. I appreciate the comments on how best to organize
the code as well. Simplifies things greatly.
Thanks,
Nate
On Wed, Sep 7, 2011 at 3:33 PM, Dennis Murphy wrote:
> Hi:
>
> For your test data, try this:
>
> # Result of dput(NerveSurv)
> NerveSu
Have a close look at ?storage.mode and ?typeof.
The problem here, as usual and as you seem to allude is: define
"integer" . Mathematically, there is an unequivocal definition, but
not so in finite precision arithmetic. Are you referring to the
machine representation (typeof()), or the approximate
Have you looked at www.r-project.org -> books? Also,
"www.r-project.org -> CRAN (select a mirror) -> Documentation:
Contributed" includes a variety of documents that are free and
potentially quite helpful.
Also, I wouldn't be shy about asking co-workers for suggestions.
(I'd ra
There was a post from Martin Maechler some years ago and I had to search a bit
to find it. For these sorts of issues, I typically trust his judgement.
The post is here:
https://stat.ethz.ch/pipermail/r-help/2003-April/032471.html
His solution also handles complex numbers.
HTH,
Marc Schwartz
Am 08.09.2011 20:48, schrieb Marc Schwartz:
> There was a post from Martin Maechler some years ago and I had to
search a bit to find it. For these sorts of issues, I typically trust
his judgement.
>
> The post is here:
>
>https://stat.ethz.ch/pipermail/r-help/2003-April/032471.html
>
> His
On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote:
Am 08.09.2011 20:48, schrieb Marc Schwartz:
> There was a post from Martin Maechler some years ago and I had to
search a bit to find it. For these sorts of issues, I typically
trust his judgement.
>
> The post is here:
>
>https://
On Sep 8, 2011, at 2:42 PM, David Winsemius wrote:
>
> On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote:
>
>> Am 08.09.2011 20:48, schrieb Marc Schwartz:
>> > There was a post from Martin Maechler some years ago and I had to search a
>> > bit to find it. For these sorts of issues, I typi
On Sep 8, 2011, at 3:54 PM, Marc Schwartz wrote:
On Sep 8, 2011, at 2:42 PM, David Winsemius wrote:
On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote:
Am 08.09.2011 20:48, schrieb Marc Schwartz:
There was a post from Martin Maechler some years ago and I had to
search a bit to find
On Thu, Sep 8, 2011 at 12:10 PM, Brian Smith wrote:
> Hi,
>
> Thanks for the reply. For the combined plot, if I use:
>
> ggplot(comb2) + geom_freqpoly(aes(x = value,
> y = ..density.., group = X2))
>
>
> I get the same colour for both the sets of distributions.
That is bec
Dear R Community,
I am new to R, and have a question that I suspect may be quite simple but is
proving a formidable roadblock for me. I have a large data set that
includes water-quality measurements collected over many 24-hour periods.
The date and time of sample collection are in a combined Dat
Dear R Group:
Based on the following data, how to do a great Spider (Radar) Plot? Any
advice is greatly appreciated.
HospID Rate Age Charlson NIHSS 1 0.2 49 3.5 0 2 0.1 48 1.8 12 3 0.4 56
2.1 5 4 0.3 77 0 7 5 0.2 67 6.5 3 6 0.1 62 4.8 4.6 7 0.1 64 12 5.2 8 0.3 61
3 2.8 9 0.15 69 4.5 1.9 1
Hi:
You could try something like this:
# rbind the two melted data frames:
comb <- rbind(mat2, tab2)
# create a group factor to distinguish the two data sets
comb <- mutate(comb, gp = factor(rep(1:2, each = 1)))
# Generate the plot:
ggplot(comb) +
geom_freqpoly(aes(x = value, y = ..dens
Try altering your subset operation from this:
with(DataSet, subset(DataSet, DataSet$NewDateTime < '2004-08-05 14:15:00'))
to this:
with(DataSet, subset(DataSet, DataSet$NewDateTime < as.POSIXct('2004-08-05
14:15:00')))
and see if you get the desired effect.
The statement DataSet$NewDateTime <
Both ??radar and ??spider return ?stars.
2011/9/8 XINLI LI :
> Dear R Group:
>
> Based on the following data, how to do a great Spider (Radar) Plot? Any
> advice is greatly appreciated.
>
> HospID Rate Age Charlson NIHSS 1 0.2 49 3.5 0 2 0.1 48 1.8 12 3 0.4 56
> 2.1 5 4 0.3 77 0 7 5 0.2 67 6
On Sep 8, 2011, at 3:09 PM, David Winsemius wrote:
>
> On Sep 8, 2011, at 3:54 PM, Marc Schwartz wrote:
>
>>
>> On Sep 8, 2011, at 2:42 PM, David Winsemius wrote:
>>
>>>
>>> On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote:
>>>
Am 08.09.2011 20:48, schrieb Marc Schwartz:
> Th
Your example works without error for me; R version 2.13.1 Patched
on Ubuntu Linux. No idea why you are having problems.
cheers,
Rolf Turner
On 09/09/11 05:24, Monsieur Do wrote:
> I've seen a number of issues with the loading of workspaces discussed
> previously, but here's
On Sep 8, 2011, at 4:41 PM, Marc Schwartz wrote:
On Sep 8, 2011, at 3:09 PM, David Winsemius wrote:
On Sep 8, 2011, at 3:54 PM, Marc Schwartz wrote:
On Sep 8, 2011, at 2:42 PM, David Winsemius wrote:
On Sep 8, 2011, at 3:35 PM, Alexander Engelhardt wrote:
Am 08.09.2011 20:48, schrie
On 09/09/11 06:46, Spencer Graves wrote:
I'd rather ask a stupid question than make a stupid mistake.
Fortune?
cheers,
Rolf Turner
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Erin,
I haven't used Rweb recently. The URL is
http://www.math.montana.edu/Rweb/ . If you have a server, you could set
up the server version of RStudio: http://rstudio.org/download/server .
It worked well when I tried it.
Best,
Matt
On Tue, 2011-09-06 at 17:07 -0500, Erin Hodgess wrote:
> Dea
Are you doing something in your profile.site file or loading a package that
masks "load"?
It seems that you're deparsing (or parsing, I can never remember which is
which) the literal results of the load function.
Does it work if you do this?
save(a, file='a.RData')
On Thu, Sep 8, 2011 at 12:24
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